1 Introduction

A generalization of the metric space can be obtained as a partial-metric space by replacing the condition \(d(x,x)=0\) with the condition \(d(x,x)\le d(x,y)\) for all x, y in the definition of the metric. In the year 1993, Czerwik [1] introduced the concept of a b-metric space as another generalization of the concept of metric space. Several authors have focused on fixed point theorems for a metric space, a partial-metric space, quasi-partial metric space and a partial b-metric space. For further information on the subject see [216].

The concept of a quasi-partial-metric space was introduced by Karapınar et al. [17]. He studied some fixed point theorems on these spaces whereas Shatanawi and Pitea [18] studied some coupled fixed point theorems on quasi-partial-metric spaces.

The aim of this paper is to introduce the concept of quasi-partial b-metric spaces which is a generalization of the concept of quasi-partial-metric spaces. The fixed point results are proved in setting of such spaces and some examples are given to verify the effectiveness of the main results.

2 Preliminaries

We begin the section with some basic definitions and concepts.

Definition 2.1

([17])

A quasi-partial metric on a non-empty set X is a function \(q:X\times X\rightarrow \mathbb{R}^{+}\), satisfying

(QPM1):

If \(q(x,x)=q(x,y)=q(y,y)\), then \(x=y\).

(QPM2):

\(q(x,x)\le q(x,y)\).

(QPM3):

\(q(x,x)\le q(y,x)\).

(QPM4):

\(q(x,y)+q(z,z)\le q(x,z)+q(z,y)\) for all \(x,y,z\in X\).

A quasi-partial-metric space is a pair \((X, q)\) such that X is a non-empty set and q is a quasi-partial metric on X.

Let q be a quasi-partial metric on the set X. Then

$$d_{q} (x,y)=q(x,y)+q(y,x)-q(x,x)-q(y,y) \mbox{ is a metric on }X. $$

Lemma 2.1

([17])

For a quasi-partial metric q on X,

$$p_{q} (x,y)=\frac{1}{2} \bigl[q(x,y)+q(y,x)\bigr] \quad\textit{for all } x,y\in X \textit{ is a partial metric on }X. $$

Lemmas 2.2

([1921])

  1. (A)

    A sequence \(\{x_{n} \} \) is Cauchy in a partial-metric space \((X,p)\) if and only if \(\{x_{n} \} \) is Cauchy in the (corresponding) metric space \((X,d_{p})\).

  2. (B)

    A partial-metric space \((X,p)\) is complete if and only if the (corresponding) metric space \((X,d_{p})\) is complete. Moreover,

    $$\lim_{n\to\infty} d_{p} (x,x_{n})=0\quad \Leftrightarrow\quad p(x,x)=\lim_{n\to\infty} p(x,x_{n})=\lim _{n,m\to\infty} p(x_{n} ,x_{m}). $$

Lemma 2.3

([17])

Let \((X,q)\) be a quasi-partial metric space, let \((X,p_{q})\) be the corresponding partial-metric space, and let \((X,d_{p_{q} })\) be the corresponding metric space. Then the following statements are equivalent:

  1. (A)

    The sequence \(\{x_{n} \} \) is Cauchy in \((X,q)\) and \((X,q)\) is complete.

  2. (B)

    The sequence \(\{x_{n} \} \) is Cauchy in \((X,p_{q})\) and \((X,p_{q})\) is complete.

  3. (C)

    The sequence \(\{x_{n} \} \) is Cauchy in \((X,d_{p_{q} })\) and \((X,d_{p_{q} })\) is complete.

Also,

$$\begin{aligned} &\lim_{n\to\infty} d_{q} (x,x_{n})=0 \quad\Leftrightarrow\quad p_{q} (x,x)=\lim_{n\to\infty} p_{q} (x,x_{n})=\lim_{n,m\to\infty}p_{q} (x_{n} ,x_{m})\\ &\hphantom{\lim_{n\to\infty} d_{q} (x,x_{n})=0}\quad\Leftrightarrow\quad q(x,x)=\lim_{n\to\infty}q(x,x_{n})= \lim_{n,m\to\infty}q(x_{n} ,x_{m})\\ &\hphantom{\lim_{n\to\infty} d_{q} (x,x_{n})=0\quad\Leftrightarrow\quad q(x,x)} =\lim_{n\to\infty} q(x_{n},x)=\lim _{n,m\to\infty} q(x_{m},x_{n}). \end{aligned}$$

Definition 2.2

([17])

If \(T:X\to X\) is any map on X, \(O(x)=\{x,Tx,T^{2} x,\ldots\}\) is called the orbit of x. A mapping \(G:X\to{\mathbb{R}}^{+} \) is T-orbitally lower semi-continuous at x if \(\{x_{n} \} \) is a sequence in \(O(x)\) and \(\lim x_{n} =z\) implies \(G(z)\le\liminf G(x_{n})\).

3 Quasi-partial b-metric space

We introduce the concept of quasi-partial b-metric space here.

Definition 3.1

A quasi-partial b-metric on a non-empty set X is a mapping \(qp_{b} :X\times X\to{\mathbb{R}}^{+}\) such that for some real number \(s \geq1\) and all \(x, y, z\in X\):

(QPb1):

\(qp_{b} (x,x)=qp_{b} (x,y)=qp_{b} (y,y)\Rightarrow x=y\),

(QPb2):

\(qp_{b} (x,x)\le qp_{b} (x,y)\),

(QPb3):

\(qp_{b} (x,x)\le qp_{b} (y,x)\),

(QPb4):

\(qp_{b} (x,y)\le s[qp_{b} (x,z)+qp_{b} (y,z)]-qp_{b} (z,z)\).

A quasi-partial b-metric space is a pair \((X,qp_{b})\) such that X is a non-empty set and \((X,qp_{b})\) is a quasi partial b-metric on X. The number s is called the coefficient of \((X,qp_{b})\).

For a quasi-partial b-metric space \((X,qp_{b})\), the function \(d_{qp_{b} } :X\times X\to{\mathbb{R}}^{+} \) defined by

$$d_{qp_{b} } (x,y)=qp_{b} (x,y)+qp_{b} (y,x)-qp_{b} (x,x)-qp_{b} (y,y) \mbox{ is a }b \mbox{-metric on }X. $$

Example 3.1

Let \(X=[0,1]\).

Define \(qp_{b}(x,y)=|x-y|+x\). Here

$$\begin{aligned} qp_{b}(x,x)=qp_{b}(x,y)=qp_{b}(y,y)\quad \Rightarrow\quad x=y \quad\mbox{as } x=|x-y|+x=y \mbox{ gives } x=y . \end{aligned}$$

Again, \(qp_{b}(x,x)\le qp_{b}(x,y)\) as \(x\le|x-y|+x\) and similarly, \(qp_{b}(x,x)\le qp_{b}(y,x)\) as \(x\le|y-x|+y\) for \(0< x<y\).

Also \(qp_{b}(x,y)+qp_{b}(z,z)\le s[qp_{b}(x,z)+qp_{b}(z,y)]\) as

$$|x-y|+x+z\le s\bigl[|x-z|+x+|z-y|+z\bigr] \quad\mbox{for all } s\ge1. $$

It can be observed that

$$\begin{aligned} |x-y|+x+z=|x-z+z-y|+x+z\le|x-z|+|z-y|+x+z . \end{aligned}$$

So \((X,qp_{b})\) is a quasi-partial b-metric space with \(s\ge1\).

Example 3.2

Let \(X=[1, \infty)\).

Define \(qp_{b} :X\times X\to{\mathbb{R}}^{+} \) as \(qp_{b} (x,y)=\ln (xy)\). Then \((X,qp_{b})\) is a quasi-partial b-metric space.

Let \(qp_{b} (x,x)=qp_{b} (x,y)=qp_{b} (y,y) \Rightarrow \ln(x^{2})=\ln(xy)=\ln(y^{2}) \Rightarrow x=y\).

Let \(x,y\in X\). Without loss of generality \(x \leq y \Rightarrow\ln x\le\ln y \Rightarrow2\ln x\le\ln x+\ln y \Rightarrow\ln(x^{2})\le\ln x+\ln y\).

Thus, \(qp_{b} (x,x)\le qp_{b} (x,y)\).

Similarly \(qp_{b} (x,x)\le qp_{b} (y,x)\).

For (QPb4) we have

$$\begin{aligned} qp_{b} (x,y) &=\ln x+\ln y \\ &\le s\ln x+s\ln y\quad\mbox{since }s\ge1\mbox{ and also }\ln x\ge0\mbox{ and }\ln y \ge0 \\ &\le s\ln x+s\ln y+2\ln z(s-1)\quad\mbox{since }\ln z\ge0\mbox{ and }s-1\ge 0 \\ &=s\bigl\{ qp_{b} (x,z)+qp_{b} (z,y)\bigr\} -qp_{b} (z,z). \end{aligned}$$

Example 3.3

Let \(X= [0, \frac{\pi}{4} ]\) and define \(qp_{b}:X\times X\to \mathbb{R}^{+}\) as

$$qp_{b} (x,y)=\sin x+\sin y. $$

Then \((X,qp_{b})\) is a quasi-partial b-metric space.

Lemma 3.4

Let \((X,qp_{b})\) be a quasi-partial b-metric space. Then the following hold:

  1. (A)

    If \(qp_{b} (x,y)=0\) then \(x = y\).

  2. (B)

    If \(x\ne y\), then \(qp_{b} (x,y)>0\) and \(qp_{b} (y,x)>0\).

The proof is similar to the case of quasi-partial-metric space [17].

Lemma 3.5

Every quasi-partial space is a quasi-partial b-metric space. But the converse does not need to be true.

Definition 3.2

Let \((X,qp_{b})\) be a quasi-partial b-metric. Then:

  1. (i)

    A sequence \(\{x_{n} \} \subset X\) converges to \(x\in X\) if and only if

    $$qp_{b} (x,x)=\lim_{n\to\infty} qp_{b} (x,x_{n})=\lim_{n\to\infty} qp_{b} (x_{n} ,x). $$
  2. (ii)

    A sequence \(\{x_{n} \} \subset X\) is called a Cauchy sequence if and only if

    $$\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m}) \quad\mbox{and}\quad \lim_{n,m\to\infty}qp_{b} (x_{m} ,x_{n}) \mbox{ exist (and are finite)}. $$
  3. (iii)

    The quasi-partial b-metric space \((X,qp_{b})\) is said to be complete if every Cauchy sequence \(\{x_{n} \}\subset X\) converges with respect to \(\tau_{qp_{b} } \) to a point \(x\in X\) such that

    $$qp_{b} (x,x)=\lim_{n,m\to\infty}qp_{b} (x_{m} ,x_{n})=\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m}). $$
  4. (iv)

    A mapping \(f:X\to X\) is said to be continuous at \(x_{0} \in X\) if, for every \(\varepsilon>0\), there exists \(\delta>0\) such that \(f(B(x_{0} ,\delta))\subset B(f(x_{0}),\varepsilon)\).

Lemma 3.6

Let \((X,qp_{b})\) be a quasi-partial b-metric space and \((X,d_{qp_{b} })\) be the corresponding b-metric space. Then \((X,d_{qp_{b} })\) is complete if \((X,qp_{b})\) is complete.

Proof

Since \((X,qp_{b})\) is complete, every Cauchy sequence \(\{x_{n} \} \) in X converges with respect to \(\tau_{qp_{b} } \) to a point \(x\in X\) such that

$$ qp_{b} (x,x)=\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m})=\lim_{n,m\to\infty} qp_{b} (x_{m} ,x_{n}). $$
(1)

Consider a Cauchy sequence \(\{x_{n} \} \) in \((X,d_{qp_{b} })\). We will show that \(\{x_{n} \} \) is Cauchy in \((X,qp_{b})\). Since \(\{x_{n} \} \) is Cauchy in \((X,d_{qp_{b} })\), \(\lim_{n,m\to\infty} d_{qp_{b} } (x_{n} ,x_{m})\) exists and is finite.

Also, \(d_{qp_{b} } (x_{n} ,x_{m})=qp_{b} (x_{n} ,x_{m})+qp_{b} (x_{m} ,x_{n})-qp_{b} (x_{n} ,x_{n})-qp_{b} (x_{m} ,x_{m})\).

Clearly, \(\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m})\) and \(\lim_{n,m\to\infty} qp_{b} (x_{m} ,x_{n})\) exist and are finite.

Therefore, \(\{x_{n} \} \) is a Cauchy sequence in \((X,qp_{b})\). Now, since \((X,qp_{b})\) is complete, the sequence \(\{x_{n} \} \) converges with respect to \(\tau_{qp_{b} } \)to a point \(x\in X\) such that (1) holds.

For \(\{x_{n} \} \) to be convergent in \((X,d_{qp_{b} })\) we will show that \(d_{qp_{b} } (x,x)=\lim_{n\to\infty} d_{qp_{b} } (x,x_{n})\).

If follows from the definition of \(d_{qp_{b} } \) that \(d_{qp_{b} } (x,x)=0\). Also,

$$\begin{aligned} \lim_{n\to\infty} d_{qp_{b} } (x,x_{n}) &=\lim _{n\to\infty} qp_{b} (x,x_{n})+\lim _{n\to\infty } qp_{b} (x_{n} ,x) -\lim _{n\to\infty} qp_{b} (x_{n},x_{n})-\lim _{n\to\infty}qp_{b} (x,x) \\ &= 0 \quad\mbox{by (1) and definition of convergence in }(X,qp_{b}). \end{aligned}$$

Hence, \(d_{qp_{b} } (x,x)=\lim_{n\to\infty} d_{qp_{b} } (x,x_{n})\). □

In [17] Karapınar et al. proved a fixed point theorem on quasi-partial-metric space. Motivated by this, we have generalized the results on a quasi-partial b-metric space.

4 The main results

Theorem 4.1

Let \((X,qp_{b})\) be a quasi-partial b-metric space, and let \(T:X\to X\). Then the following hold:

  1. (A)

    There exists \(\phi:X\to{\mathbb{R}}^{+} \) such that

    $$\begin{aligned} &qp_{b} (x,Tx)\le\phi(x)-\phi(Tx) \quad\textit{for all } x \in X \quad\textit{if and only if} \\ &\quad\sum_{n=0}^{\infty} qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr) \textit{ converges for all } x\in X. \end{aligned}$$
  2. (B)

    There exists \(\phi:X\to{\mathbb{R}}^{+} \) such that

    $$\begin{aligned} &qp_{b} (x,Tx)\le\phi(x)-\phi(Tx) \quad\textit{for all }x\in O(x)\quad\textit{if and only if} \\ &\quad\sum_{n=0}^{\infty} qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr) \textit{ converges for all }x\in O(x). \end{aligned}$$

Proof

(A) Let \(x\in X\), and let

$$qp_{b} (x,Tx)\le\phi(x)-\phi(Tx). $$

Define the sequence \(\{x_{n} \} _{n=1}^{\infty} \) in the following way:

$$x_{0} =x \quad\mbox{and} \quad x_{n+1} =Tx_{n} =T^{n+1} x_{0}, \quad\mbox{for all } n = 0, 1, 2, \ldots. $$

Set \(z_{n} (x)=\sum_{k=0}^{n} qp_{b} (x_{k} ,x_{k+1})=\sum_{k=0}^{n} qp_{b} (T^{k} x_{0} ,T^{k+1} x_{0})\). Then

$$\begin{aligned} z_{n} (x) &\le\sum_{k=0}^{n} \bigl[\phi\bigl(T^{k} x_{0}\bigr)-\phi\bigl(T^{k+1} x_{0}\bigr)\bigr] \\ &=\bigl[\phi(x_{0})-\phi(Tx_{0})\bigr]+\cdots+\bigl[\phi \bigl(T^{n} x_{0}\bigr)-\phi\bigl(T^{n+1} x_{0}\bigr)\bigr] \\ &=\bigl[\phi(x_{0})-\phi\bigl(T^{n+1} x_{0}\bigr) \bigr]\le\phi(x_{0})=\phi(x). \end{aligned}$$
(2)

Thus, (2) implies that \(\{z_{n} (x)\} \) is bounded. Also \(\{ z_{n} (x)\} \) is non-decreasing and hence convergent. Therefore, \(\sum_{n=0}^{\infty} qp_{b} (T^{n} x,T^{n+1} x)\) converges.

Conversely, define

$$\phi(x)=\sum_{n=0}^{\infty} qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr) \quad\mbox{and}\quad z_{n} (x)=\sum_{k=0}^{n} qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr). $$

Then

$$\phi(Tx)=\sum_{n=0}^{\infty} qp_{b} \bigl(T^{n+1} x,T^{n+2} x\bigr) \quad\mbox{and}\quad z_{n} (Tx)=\sum_{k=0}^{n} qp_{b} \bigl(T^{k+1} x,T^{k+2} x\bigr). $$

Using these definitions, we get

$$ \begin{aligned}[b] z_{n} (x)-z_{n} (Tx) &=\sum _{k=0}^{n} qp_{b} \bigl(T^{k} x,T^{k+1} x\bigr)-\sum_{k=0}^{n} qp_{b} \bigl(T^{k+1} x,T^{k+2} x\bigr)\\ &=qp_{b} (x,Tx)-qp_{b} \bigl(T^{n+1} x,T^{n+2} x\bigr) . \end{aligned} $$
(3)

Since \(\sum_{n=0}^{\infty} qp_{b} (T^{n} x, T^{n+1} x)\) converges for all \(x \in X\),

$$\lim_{n\to\infty}z_{n} (x)=\phi(x) \quad\mbox{and}\quad \lim_{n\to\infty}qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr)=0. $$

Letting \(n\to\infty\) in (3) gives \(qp_{b} (x,Tx)=\phi (x)-\phi(Tx)\).

(B) It can easily be proved using part (A). □

Example 4.1

Let \(X=[0, 1]\). Define \(qp_{b} (x,y)=|x-y| + |x|\).

Then \(qp_{b} (x,y)\) satisfies all conditions of quasi-partial b-metric space. It is also quasi-partial metric. But for \(x\ne y\), \(qp_{b} (x,y)\ne qp_{b} (y,x)\) and \(qp_{b} (x,x)\ne0\) for \(x\ne0\). So \(qp_{b} \) is not a partial metric or a quasi-metric. Define \(T:X\to X\) as \(Tx=\frac{x}{3} \) for all \(x\in X\). Then the series \(\sum_{n=0}^{\infty} qp_{b} (T^{n} x,T^{n+1} x)\) is convergent. Indeed,

$$\begin{aligned} \sum_{n=0}^{\infty} qp_{b} \bigl(T^{n} x,T^{n+1} x\bigr) &=\sum _{n=0}^{\infty} qp_{b} \biggl(\frac{x}{3^{n} }, \frac {x}{3^{n+1} } \biggr) =\sum_{n=0}^{\infty} \biggl\vert \frac {x}{3^{n} } -\frac{x}{3^{n+1} } \biggr\vert +\biggl\vert \frac{x}{3^{n} } \biggr\vert \\ &=\sum_{n=0}^{\infty} \biggl\vert \frac{2x}{3^{n+1} } \biggr\vert +\biggl\vert \frac{x}{3^{n} } \biggr\vert = \sum_{n=0}^{\infty} \frac {5x}{3^{n+1} } = \frac{5x}{3}\cdot\frac{1}{1-\frac{1}{3} } =\frac {5x}{2} . \end{aligned}$$

Then the conditions of Theorem 4.1 are satisfied for \(\phi(x)=\frac {5x}{2} \). Indeed

$$\begin{aligned} qp_{b} (x,Tx) &=qp_{b} \biggl(x,\frac{x}{3} \biggr)= \biggl\vert x-\frac{x}{3} \biggr\vert +|x|=\biggl\vert \frac{2x}{3} \biggr\vert +|x|=\frac{5x}{3}=\phi(x)-\phi(Tx). \end{aligned}$$

The next result gives conditions for the existence of fixed points of operators on quasi-partial b-metric space.

Theorem 4.2

Let \((X, qp_{b})\) and \((Y, qp_{b})\) be complete quasi-partial b-metric spaces. Let also \(T:X\to X\), \(R: X\to Y\), and \(\phi:R(X)\to{\mathbb{R}}^{+} \). If there exist \(x\in X\) and \(c>0\) such that

$$ \max\bigl\{ qp_{b} (y, Ty), cqp_{b} (Ry, RTy)\bigr\} \le\phi(Ry)-\phi(RTy) $$
(4)

for all \(y\in O(x)\), then the following hold:

  1. (A)

    \(\lim_{n\to\infty} T^{n} x=z\) exists.

  2. (B)

    \(Tz=z\) if and only if \(G(x)=qp_{b} (x, Tx)\) is T-orbitally lower semi-continuous at x.

  3. (C)

    \(qp_{b} (x, T^{n} x)\le s^{n-1} \phi(Rx)\).

  4. (D)

    For \(m > n\), \(qp_{b} (T^{n} x, T^{m} x)\le s^{m-n} [\phi(RT^{n} x)]\).

Proof

(A) Let \(x\in X\). Define the sequence \(\{x_{n} \} _{n=1}^{\infty} \) as follows:

$$x_{0} =x \quad\mbox{and}\quad x_{n+1} =Tx_{n} =T^{n+1} x_{0}, \quad\mbox{for all } n = 0, 1, 2,\ldots . $$

We will show that \(\{x_{n} \} _{n=1}^{\infty} \) is Cauchy.

Using (QPb4), we get

$$\begin{aligned} qp_{b} (x_{n} ,x_{n+2}) &\le s\bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} -qp_{b} (x_{n+1},x_{n+1}) \\ &\le s\bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} \end{aligned}$$
(5)

and, similarly,

$$\begin{aligned} qp_{b} (x_{n},x_{n+3}) &\le s\bigl\{ qp_{b} (x_{n},x_{n+2})+qp_{b} (x_{n+2},x_{n+3})\bigr\} -qp_{b} (x_{n+2},x_{n+2}) \\ &\le s^{2} \bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} + s\bigl\{ qp_{b} (x_{n+2},x_{n+3})\bigr\} . \end{aligned}$$
(6)

Now,

$$\begin{aligned} qp_{b} (x_{n},x_{n+4}) \le{}& s\bigl\{ qp_{b} (x_{n},x_{n+3})+qp_{b} (x_{n+3},x_{n+4})\bigr\} -qp_{b} (x_{n+3},x_{n+3}) \\ \ \le{}& s^{3} \bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} +s^{2} \bigl\{ qp_{b} (x_{n+2},x_{n+3})\bigr\} \\ &{} +s\bigl\{ qp_{b} (x_{n+3},x_{n+4})\bigr\} . \end{aligned}$$

On generalization, we get

$$\begin{aligned} qp_{b} (x_{n},x_{m}) \le{}& s^{m-n-1} \bigl\{ qp_{b} (x_{n},x_{n+1})+qp_{b} (x_{n+1},x_{n+2})\bigr\} \\ &{} +s^{m-n-2} \bigl\{ qp_{b} (x_{n+2},x_{n+3}) \bigr\} +\cdots+s\bigl\{ qp_{b} (x_{m-1},x_{m})\bigr\} \\ \le{}& s^{m-n-1} \bigl\{ qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)+qp_{b} \bigl(T^{n+1} x, T^{n+2} x\bigr)\bigr\} \\ &{} +s^{m-n-2} \bigl\{ qp_{b} \bigl(T^{n+2} x, T^{n+3} x\bigr)\bigr\} +\cdots+s\bigl\{ qp_{b} \bigl(T^{m-1} x, T^{m} x\bigr)\bigr\} \\ ={}&\sum_{k=n+1}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} +s^{m-n-1} qp_{b} (x_{n},x_{n+1}) \\ ={}&\sum_{k=n}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} +s^{m-n-1} qp_{b} (x_{n},x_{n+1}) -s^{m-n} qp_{b} (x_{n},x_{n+1}) \\ ={}&\sum_{k=n}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} -s^{m-n} qp_{b} (x_{n},x_{n+1}) \biggl[1- \frac{1}{s} \biggr] \\ \le{}&\sum_{k=n}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} \quad \mbox{for } m > n. \end{aligned}$$
(7)

Set \(z_{n} (x)=\sum_{k=0}^{n} s^{m-k} \{qp_{b} (T^{k} x, T^{k+1} x)\} \).

From (4) we have

$$\begin{aligned}& \begin{aligned}[b] s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} &\le s^{m-k} \max\bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr), cqp_{b} \bigl(RT^{k} x, RT^{k+1} x\bigr)\bigr\} \\ & \le s^{m-k} \bigl\{ \phi\bigl(RT^{k} x\bigr)-\phi \bigl(RT^{k+1} x\bigr)\bigr\} \quad\mbox{for all } k = 0, 1, \ldots \end{aligned} \end{aligned}$$
(8)
$$\begin{aligned}& \begin{aligned}[b] \quad\Rightarrow\quad z_{n} (x)\le{}&\sum _{k=0}^{n} s^{m-k} \bigl\{ \phi \bigl(RT^{k} x\bigr)-\phi\bigl(RT^{k+1} x\bigr)\bigr\} \\ \le{}& s^{m} \phi(Rx)-s^{m} \phi(RTx)+s^{m} \phi(RTx)-s^{m-1} \phi \bigl(RT^{2} x\bigr)+\cdots \\ &{} +s^{m-n+1} \phi\bigl(RT^{n} x\bigr)-s^{m-n} \phi \bigl(RT^{n+1} x\bigr) \\ ={}&s^{m} \phi(Rx)-s^{m-n} \phi\bigl(RT^{n+1} x \bigr) \\ \le{}& s^{m} \phi(Rx). \end{aligned} \end{aligned}$$
(9)

Thus, \(\sum_{k=0}^{\infty} s^{m-k} \{ qp_{b} (T^{k} x, T^{k+1} x)\}\) is convergent.

$$\begin{aligned} \Rightarrow \quad\sum_{n=0}^{\infty} s^{m-n} \bigl\{ qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)\bigr\} \mbox{ is convergent}. \end{aligned}$$

Taking the limit as \(n,m\to\infty\) in (7), we get

$$ \lim_{m,n\to\infty} qp_{b} (x_{n},x_{m})=\lim_{m,n\to \infty} \bigl(z_{m-1} (x)-z_{n-1} (x)\bigr)=0. $$
(10)

Using similar arguments,

$$\begin{aligned} \lim_{m,n\to\infty} qp_{b} (x_{m} ,x_{n})=0. \end{aligned}$$
(11)

Thus the sequence \(\{x_{n} \} \) is Cauchy in \((X,qp_{b})\). Since \((X,qp_{b})\)is complete, \((X,d_{qp_{b} })\) is also complete by Lemma 2.3, and hence \(\lim_{n\to\infty} d_{qp_{b} } (T^{n} x,z)=0\), \(\lim_{n\to\infty} T^{n} x=z\).

Further, \(\lim_{n\to\infty} qp_{b} (T^{n} x, T^{n+1} x)=0\) and hence \(\lim_{n\to\infty} qp_{b} (T^{n} x, T^{n+1} x)=qp_{b} (z,z)=0\).

(B) Assume that \(Tz=z\) and that \(x_{n} \) is a sequence in \(O(x)\) with \(x_{n} \to z\).

By Lemma 3.6,

$$\begin{aligned} \lim_{n\to\infty} d_{qp_{b} } (z,x_{n})=0 \quad \Leftrightarrow\quad qp_{b} (z,z)=\lim_{n\to\infty} qp_{b} (z,x_{n}) =\lim_{n,m\to\infty} qp_{b} (x_{n} ,x_{m}). \end{aligned}$$
(12)

Then \(G(z)=qp_{b} (z,Tz)=qp_{b} (z,z)\le\lim_{n\to\infty} \inf qp_{b} (x_{n} ,Tx_{n})=\lim_{n\to\infty} \inf G(x_{n})\).

Thus G is T-orbitally lower semi-continuous at x.

Conversely, suppose that \(x_{n} =T^{n} x\to z\) and that G is T-orbitally lower semi-continuous at x. Then

$$\begin{aligned} 0&\le qp_{b} (z, Tz)=G(z) \le\lim_{n\to\infty} \inf G(x_{n})=\lim_{n\to\infty } \inf qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr) \\ &=\lim_{n\to\infty} \inf qp_{b} (x_{n},x_{n+1})= qp_{b} (z,z)=0. \end{aligned}$$
(13)

By Lemma 3.4, we have \(Tz=z\).

(C) We have, from (QPb4) and (4),

$$\begin{aligned}& \begin{aligned}[b] qp_{b} \bigl(x, T^{2} x\bigr) &\le s \bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x \bigr)\bigr\} -qp_{b} (Tx, Tx) \\ &\le s\bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} , \end{aligned} \\& \begin{aligned}[b] qp_{b} \bigl(x, T^{3} x\bigr)&\le s \bigl\{ qp_{b} \bigl(x, T^{2} x\bigr)+qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr\} -qp_{b} \bigl(T^{2} x, T^{2} x\bigr) \\ &\le s\bigl[s\bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} +qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr] \\ &\le s^{2} \bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} +s\bigl\{ qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr\} . \end{aligned} \end{aligned}$$

On generalization, we get

$$\begin{aligned} &qp_{b} \bigl(x, T^{n} x\bigr) \\ &\quad\le s^{n-1} \bigl\{ qp_{b} (x, Tx)+qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} +s^{n-2} \bigl\{ qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr\} + \cdots \\ &\qquad{} +s\bigl\{ qp_{b} \bigl(T^{n-1} x, T^{n} x \bigr)\bigr\} \\ &\quad\le s^{n-1} \bigl\{ qp_{b} (x, Tx)\bigr\} +s^{n-1} \bigl\{ qp_{b} \bigl(Tx, T^{2} x\bigr)\bigr\} +s^{n-2} \bigl\{ qp_{b} \bigl(T^{2} x, T^{3} x\bigr)\bigr\} +\cdots \\ &\qquad{} +s\bigl\{ qp_{b} \bigl(T^{n-1} x, T^{n} x \bigr)\bigr\} \\ &\quad\le s^{n-1} \bigl\{ \phi(Rx)-\phi(RTx)\bigr\} +s^{n-1} \bigl\{ \phi(RTx)-\phi \bigl(RT^{2} x\bigr)\bigr\} \\ &\qquad{} +s^{n-2} \bigl\{ \phi\bigl(RT^{2} x\bigr)-\phi \bigl(RT^{3} x\bigr)\bigr\} +\cdots+s\bigl\{ \phi \bigl(RT^{n-1} x\bigr)-\phi\bigl(RT^{n} x\bigr)\bigr\} \\ &\quad\le s^{n-1} \phi(Rx)-s^{n-1} \phi\bigl(RT^{2} x\bigr)+s^{n-2} \phi\bigl(RT^{2} x\bigr) -s^{n-2} \phi\bigl(RT^{3} x\bigr)+\cdots \\ &\qquad{} +s\phi\bigl(RT^{n-1} x\bigr)-s\phi\bigl(RT^{n} x \bigr) \\ &\quad\le s^{n-1} \phi(Rx)-s\phi\bigl(RT^{2} x\bigr)-s\phi \bigl(RT^{n-1} x\bigr)-s\phi \bigl(RT^{n} x\bigr) \\ &\quad\le s^{n-1} \phi(Rx). \end{aligned}$$
(14)

(D) From (7) we get

$$qp_{b} (x_{n},x_{m})\le\sum _{k=n}^{m-1} s^{m-k} \bigl\{ qp_{b} \bigl(T^{k} x, T^{k+1} x\bigr)\bigr\} \quad\mbox{for } m >n. $$

Note that

$$\begin{aligned} &\sum_{k=n}^{m-1} s^{m-k} qp_{b} \bigl(T^{k} x, T^{k+1} x \bigr) \\ &\quad\le\sum_{k=n}^{m-1} s^{m-k} \bigl[\phi\bigl(RT^{k} x\bigr)-\phi\bigl(RT^{k+1} x\bigr)\bigr] \\ &\quad =s^{m-n} \phi\bigl(RT^{n} x\bigr)-s^{m-n} \phi\bigl(RT^{n+1} x\bigr)+s^{m-n-1} \phi\bigl(RT^{n+1} x \bigr) \\ &\qquad{} -s^{m-n-1} \phi\bigl(RT^{n+2} x\bigr)+\cdots+s\phi \bigl(RT^{m-1} x\bigr)-s\phi\bigl(RT^{m} x\bigr) \\ &\quad =s^{m-n} \phi\bigl(RT^{n} x\bigr)-s\phi \bigl(RT^{n+1} x\bigr)-s\phi\bigl(RT^{m-1} x\bigr)-s\phi \bigl(RT^{m} x\bigr) \\ & \quad\le s^{m-n} \phi\bigl(RT^{n} x\bigr). \end{aligned}$$
(15)

Here, \(0\le qp_{b} (x_{n},x_{m}) =qp_{b} (T^{n} x, T^{m} x)\le s^{m-n} \phi(RT^{n} x)\) for \(m>n\). □

Example 4.2

Let \(X=Y=[0,1]\). Define \(qp_{b}(x,y)=|x-y|+x\). Then \(qp_{b}\) is a quasi-partial b-metric with \(s=1\). Also define \(T:X\to X\) as \(T(x)=\frac{x}{3}\); \(R:X\to Y\) as \(R(x)=3x\), and \(\phi:R(X)\to \mathbb{R}^{+}\) as \(\phi(x)=3x\). Then for \(c=1\) and \(x\in [0,1]\) we have

$$\begin{aligned} \max\bigl\{ qp_{b}(y,Ty),cqp_{b}(Ry,RTy)\bigr\} &=\max \biggl\{ qp_{b} \biggl(y,\frac{y}{3} \biggr),qp_{b}(3y,y) \biggr\} \\ &=\max \biggl\{ \biggl\vert y-\frac{y}{3}\biggr\vert +y,|3y-y|+3y \biggr\} \\ &=\max \biggl\{ \frac{5y}{3},5y \biggr\} =5y< 6y=\phi(3y)-\phi(y) \\ &=\phi(Ry)-\phi(RTy). \end{aligned}$$

We now prove that (A), (B), (C), and (D) of the above theorem hold:

(A) \(\lim_{n\to\infty}T^{n}x=\lim_{n\to\infty}\frac {x}{3^{n}}=0=z\) (say).

So \(\lim_{n\to\infty}T^{n}x=z\) exists.

(B) By (A) part above, \(z=0\).

Therefore \(T(z)=T(0)=0=z\) holds trivially.

Hence whenever \(G(x)=qp_{b}(x,Tx)\) is T-orbitally lower semi-continuous at x then \(Tz=z\).

Conversely, let \(Tz=z\) and we show that G is T-orbitally lower semi-continuous at x, i.e.,

$$G(z)\le\lim\inf G(x_{n}) \quad\forall \{x_{n}\}\subseteq O(x), x_{n}\to z. $$

Let \(\{x_{n}\}\subseteq O(x)\) be a sequence converging to z. Then

$$\begin{aligned} G(z) &=qp_{b}(z,Tz)=qp_{b}(z,z)=z \\ &=\frac{5z}{3} \ (\mbox{as }z=0) =\lim\inf\frac{5x_{n}}{3} \\ &=\lim\inf\frac{2x_{n}}{3}+x_{n}=\lim\inf\biggl\vert x_{n}-\frac{x_{n}}{3}\biggr\vert +x_{n} \\ &=\lim\inf qp_{b} \biggl(x_{n},\frac{x_{n}}{3} \biggr)= \lim\inf qp_{b} (x_{n},Tx_{n})=\lim\inf G(x_{n}) . \end{aligned}$$

Hence \(G(z)=\lim\inf G(x_{n})\).

\(\begin{array}{ll} (\mathrm{C})\ \displaystyle qp_{b}\bigl(x,T^{n}x\bigr) &\displaystyle=qp_{b} \biggl(x, \frac{x}{3^{n}} \biggr)=\biggl \vert x-\frac{x}{3^{n}}\biggr \vert +x=x \biggl(2-\frac{1}{3^{n}} \biggr)< x(9) \quad \forall n\in N\\ \displaystyle&\displaystyle=\phi(3x) =s^{n-1}\phi(Rx) \quad\mbox{where } s=1. \end{array}\)

(D) Let \(m>n\) then

$$\begin{aligned} qp_{b}\bigl(T^{n}x,T^{m}x\bigr) &=qp_{b} \biggl(\frac{x}{3^{n}},\frac{x}{3^{m}} \biggr)=\biggl\vert \frac {x}{3^{n}}-\frac{x}{3^{m}}\biggr\vert +\frac{x}{3^{n}} \\ &=\frac{x}{3^{n}} \biggl[2-\frac{1}{3^{m-n}} \biggr]< \frac{x}{3^{n}}(9) \quad \forall n\in N \\ &=\phi \biggl(\frac{x}{3^{n-1}} \biggr)=\phi\bigl(3T^{n} x \bigr)=s^{m-n}\bigl[\phi\bigl(RT^{n} x\bigr)\bigr] \quad \mbox{where }s=1. \end{aligned}$$

Corollary 4.3

Let \((X, qp_{b})\) be a complete quasi-partial b-metric space. Let \(T:X\to X\) and \(\phi:X\to{\mathbb{R}}^{+} \). Suppose that there exists \(x\in X\) such that

$$qp_{b} (y, Ty)\le\phi(y)-\phi(Ty) \quad\textit{for all } y\in O(x). $$

Then the following hold:

  1. (A)

    \(\lim_{n\to\infty} T^{n} x=z\) exists.

  2. (B)

    \(Tz=z\) if and only if \(G(x)=qp_{b} (x, Tx)\) is T-orbitally lower semi-continuous at x.

  3. (C)

    \(qp_{b} (x, T^{n} x)\le s^{n-1} \phi(x)\).

  4. (D)

    For \(m > n\), \(qp_{b} (T^{n} x, T^{m} x)\le s^{m-n} \phi(T^{n} x)\).

Proof

Take \(Y = X\), \(R = I\), and \(c = 1\) in Theorem 4.2. □

Corollary 4.4

Let \((X, qp_{b})\) be a complete quasi-partial b-metric space, and let \(0< k<1\). Suppose that \(T:X\to X\) and that there exists \(x\in X\) such that

$$\begin{aligned} qp_{b} \bigl(Ty, T^{2} y\bigr)\le kqp_{b} (y, Ty) \quad\textit{for all } y\in O(x). \end{aligned}$$
(16)

Then the following hold:

  1. (A)

    \(\lim_{n\to\infty} T^{n} x=z\) exists.

  2. (B)

    \(Tz=z\) if and only if \(G(x)=qp_{b} (x, Tx)\) is T-orbitally lower semi-continuous at x.

  3. (C)

    \(qp_{b} (x, T^{n} x)\le\frac{s^{n-1} }{1-k} qp_{b} (x, Tx)\).

Proof

Set \(\phi(y)=\frac{1}{1-k} qp_{b} (y, Ty)\) for \(y\in O(x)\).

Let \(y=T^{n} x\) in (16). Then

$$qp_{b} \bigl(T^{n+1} x, T^{n+2} x\bigr)\le kqp_{b} \bigl(T^{n} x, T^{n+1} x\bigr) $$

and

$$qp_{b} \bigl(T^{n} x, T^{n+1} x \bigr)-kqp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)\le qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)-qp_{b} \bigl(T^{n+1} x, T^{n+2} x\bigr). $$

Thus, \(qp_{b} (T^{n} x, T^{n+1} x)\le\frac{1}{1-k} [qp_{b} (T^{n} x, T^{n+1} x)-qp_{b} (T^{n+1} x, T^{n+2} x)]\) or \(qp_{b} (y,Ty)\le[\phi(y)-\phi(Ty)]\).

(A)-(C) follow immediately from Corollary 4.3. □

Corollary 4.5

Let \((X,qp_{b})\) be a complete quasi-partial b-metric space where \(qp_{b} \) is continuous. Let \(T:X\to X\) and \(\phi:X\to{\mathbb{R}}^{+} \) is continuous. Suppose that there exists \(x\in X\) such that

$$qp_{b} (y, Ty)\le\phi(y)-\phi(Ty) \quad\textit{for all } y\in O(x). $$

Then the following hold:

  1. (A)

    \(\lim_{n\to\infty} T^{n} x=z\) exists.

  2. (B)

    \(q_{p} (z,z)\le s\phi(z)\).

Proof

In Theorem 4.2(D) taking \(m = n + 1\), \(R = I\), \(c = 1\), and \(Y = X\),

$$qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)\le s \bigl[ \phi\bigl(RT^{n} x\bigr)\bigr]. $$

Now taking \(\lim n\to\infty\)

$$\begin{aligned} &\lim_{n\to\infty} qp_{b} \bigl(T^{n} x, T^{n+1} x\bigr)\le\lim_{n\to\infty} s \bigl[\phi \bigl(T^{n} x\bigr)\bigr], \\ &qp_{b} (z,z)\le s\phi(z). \end{aligned}$$

 □